If the foci of the ellipse frac{x^{2}}{16}+fr | vedclass

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(D) The given hyperbola is $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$.Rewriting in standard form: $\frac{x^{2}}{144/25} - \frac{y^{2}}{81/25} =

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$7$

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(D) The given hyperbola is $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$.Rewriting in standard form: $\frac{x^{2}}{144/25} - \frac{y^{2}}{81/25} = 1$.Here,$a^{2} = \frac{144}{25}$ and $b^{2} = \frac{81}{25}$.The eccentricity $e$ of the hyperbola is $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.The foci of the hyperbola are $(\pm ae, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$,we have $a^{2} = 16$. Let its eccentricity be $e'$.The foci of the ellipse are $(\pm ae', 0) = (\pm 4e', 0)$.Since the foci coincide,$4e' = 3$,so $e' = \frac{3}{4}$.Using the relation $e'^{2} = 1 - \frac{b^{2}}{a^{2}}$ for the ellipse:$(\frac{3}{4})^{2} = 1 - \frac{b^{2}}{16} \Rightarrow \frac{9}{16} = 1 - \frac{b^{2}}{16}$.$\frac{b^{2}}{16} = 1 - \frac{9}{16} = \frac{7}{16}$.Therefore,$b^{2} = 7$.

If the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ $(b^{2} < 16)$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide,then the value of $b^{2}$ is

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